A Piston/ cylinder of 0.5 m^3 contains water with quality 50% at 200 kPa. Slow e

Question

A Piston/ cylinder of 0.5 m^3 contains water with quality 50% at 200 kPa. Slow expansion is performed while there is heat transfer and the water is at constant pressure. The process stops when the volume has tripled. Solve for the rest questions with correct units. analysis: a) write out the general energy equation. b) simplify the given problem a) calculate or find the specific volume, temperature, and the specific internal energy for the initial state 1. b) Determine the state (compressed liquid, saturated vapor, superheated vapor, ect.) after the expansion. c) Find the temperature and specific internal energy for the final state 2. d) Calculate the work and the heat transfer. e) indicate the process in the following diagrams with labeled temperature (T1, T2), volumes (V1, V2) and pressure (P) on the curves/lines and axis

Explanation / Answer

at 200 kpa and 50 % quality :

specific volume= vf + x. vfg

= 0.00106 + 0.5 * 0.884 = 0.443 m^3/Kg

so mass of water = V/v = 0.5/0.443 = 1.1285 kg

nd also enthalpy at state 1 , h1 = hf + x.hfg = 504.7 + 0.5 * 2201.6 = 1605.5 KJ/Kg

now at state 2 : pressure = P2 as the process is solw

and v2 = 3v1 = 3 * 0.443 = 1.328 m^3/Kg

so at P2 = 200 Kpa and v2 = 1.328 m^3/kg. from steam table

enthalpy = h2 = 3075 KJ/Kg

a) enrgy equaiton :

m.h1 + W + Q o ut= m.h2

b) already simplified above

c) specific volume : v1 = 0.443 m^3/kg

and v2 = 1.328 m^3/Kg

temperature : T1 = saturation temp at 200 Kpa = 120 C

and T2 = 300 C, from steam table .

internal enrgy = u = h + pv

so, u1 = 1903 KJ/Kg

u2 = 3103 KJ/Kg

d) Work = P.dV = 200 x (3*0.5 - 0.5) = 200 KJ

NOw : m.h1 + W + Q o ut= m.h2

so, 1.1285 * 1605.5 + 200 + Q out = 1.1285 * 3075

so, Q out =1458.3 KJ

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